Simpler proof Here is an argument along the line of Andy's that works for arbitrary groups or even algebras that avoids the radical and just uses Burnside's theorem that a finite dimensional representation of a $K$-algebra over an algebraically closed field $K$ is irreducible if and only if it is surjective.

Let $A,B$ be $K$-algebras (not necessarily finite dimensional) with $K$-algebraically closed (they could be group algebras) and let $W$ be a finite dimensional irreducible $A\otimes B$-module. Let $U$ be an irreducible $A$-subrepresentation of $W$. Then the sum $S$ of all irreducible $A$-submodules of $W$ isomorphic to $U$ is invariant under any $A$-enomorphism of $W$ and hence under $B$ as the $B$-action commutes with $A$. Hence $S$ is $A\otimes B$-invariant and so $S=W$ by irreducibility. Thus $W\cong U^m$ for some $m$ as an $A$-module (this is a standard argument). Thus, up to isomorphism, we may assume that $W=U\otimes V$ with $V$ a vector space of dimension $m$ and where $A$ acts via matrices of the form $\rho(a)\otimes 1$ where $\rho$ is the irreducible representation associated to $U$. By Burnside, $\rho$ is onto $End_k(U)$. But since $End_K(U\otimes V)= End_k(U)\otimes End_k(V)$, it follows that the centralizer of $End_K(U)\otimes 1$ in $End_K(W)$ is $1\otimes End_k(V)$. Since the action of $B$ commutes with that of $A$, as a representation of $B$, we have $W$ is of the form $b\mapsto 1\otimes \psi(b)$ for some representation $\psi$ of $B$ on $V$. Clearly, if $V$ is not irreducible, then neither is $U\otimes V$ and so we are done.

If your groups are finite, then Andy’s answer is perfectly fine. If you want to do topological groups you must be slightly (but not much more) careful. First since we are dealing with finite dimensional representations, algebraic and topological reducibility are the same.

It is enough to prove that your irreducible representation decomposes as a tensor product in the case of discrete groups. For if it is isomorphic to one of the form $U\otimes V$, then since the linear isomorphism is continuous, it is enough to observe that restricting the original representation to $G$ gives $\dim V$ copies of $U$ and restricting to $H$ gives $\dim U$ copies of $V$ and so $U$ and $V$ must give continuous representations of $G,H$.

So we may assume that $G,H$ are discrete and we are dealing with no topology. Then replacing $G,H$ with their group algebras and $\mathbb C$ by algebras over an algebraically closed field, it suffices to show that if $A,B$ are $K$-algebras with $K$ algebraically closed, then any finite dimensional irreducible representation of $A\otimes B$ is equivalent to a tensor product of irreducible representations of $A$ and $B$.

But the image of $A$ and $B$ under any finite dimensional representation is a finite dimensional algebra and so without loss of generality we may assume that $A,B$ are finite dimensional.

The crux of the matter, which is basically Andy’s proof in different words, is the special case that $A,B$ are matrix algebras over $K$. Then $M_n(K)\otimes M_m(K)\cong M_{nm}(K)$ and the isomorphism intertwines the actions on $K^n\otimes K^m\cong K^{nm}$. Thus in the case $A,B$ are matrix algebras then the unique $A\otimes B$ irreducible rep is the tensor product of the unique irreducible $A$ and $B$ reps. Since arbitrary semisimple algebras over an algebraically closed field are finite direct products of matrix algebras and tensor product distributes over direct product, this handles the case $A,B$ are semisimple and also shows that $A\otimes B$ is semisimple in this case.

If $A,B$ are general finite dimensional $K$-algebras with $K$-algebraically closed, then $A/rad(A)\otimes B/rad(B)$ is semisimple by the above. Moreover, the kernel of the natural map $A\otimes B\to A/rad(A)\otimes B/rad(B)$ is easily checked to be $A\otimes rad(B)+rad(A)\otimes B$, which is a nilpotent ideal. Thus $(A\otimes B)/rad(A\otimes B)\cong A/rad(A)\otimes B/rad(B)$ and so the desired result that irreducible reps are tensor products follows since $C$ and $C/rad(C)$ have the same irreducible reps for any $K$-algebra $C$.

If your groups are finite, then Andy’s answer is perfectly fine. If you want to do topological groups you must be slightly (but not much) more careful. First since we are dealing with finite dimensional representations, algebraic and topological reducibility are the same.

It is enough to prove that your irreducible representation decomposes as a tensor product in the case of discrete groups. For if it is isomorphic to one of the form $U\otimes V$, then since the linear isomorphism is continuous, it is enough to observe that restricting the original representation to $G$ gives $\dim V$ copies of $U$ and restricting to $H$ gives $\dim U$ copies of $V$ and so $U$ and $V$ must give continuous representations of $G$, $H$.

So we may assume that $G$, $H$ are discrete and we are dealing with no topology. Then replacing $G$, $H$ with their group algebras and $\mathbb C$ by algebras over an algebraically closed field, it suffices to show that if $A$, $B$ are $K$-algebras with $K$ algebraically closed, then any finite dimensional irreducible representation of $A\otimes B$ is equivalent to a tensor product of irreducible representations of $A$ and $B$.

But the image of $A$ and $B$ under any finite dimensional representation is a finite dimensional algebra and so without loss of generality we may assume that $A$, $B$ are finite dimensional.

The crux of the matter, which is basically Andy’s proof in different words, is the special case that $A,B$ are matrix algebras over $K$. Then $M_n(K)\otimes M_m(K)\cong M_{nm}(K)$ and the isomorphism intertwines the actions on $K^n\otimes K^m\cong K^{nm}$. Thus in the case $A$, $B$ are matrix algebras then the unique $A\otimes B$ irreducible rep is the tensor product of the unique irreducible $A$ and $B$ reps. Since arbitrary semisimple algebras over an algebraically closed field are finite direct products of matrix algebras and tensor product distributes over direct product, this handles the case $A$, $B$ are semisimple and also shows that $A\otimes B$ is semisimple in this case.

$\DeclareMathOperator\rad{rad}$If $A$, $B$ are general finite dimensional $K$-algebras with $K$-algebraically closed, then $A/{\rad(A)}\otimes B/{\rad(B)}$ is semisimple by the above. Moreover, the kernel of the natural map $A\otimes B\to A/{\rad(A)}\otimes B/{\rad(B)}$ is easily checked to be $A\otimes \rad(B)+\rad(A)\otimes B$, which is a nilpotent ideal. Thus $(A\otimes B)/{\rad(A\otimes B)}\cong A/{\rad(A)}\otimes B/{\rad(B)}$ and so the desired result that irreducible reps are tensor products follows since $C$ and $C/{\rad(C)}$ have the same irreducible reps for any $K$-algebra $C$.